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3 years ago

15

What number am I? My tens digit is double my ones digit. I am less than 70 and greater than 60

1 answer:

sertanlavr [38]3 years ago

8 0

65 because it’s right in the middles of 70 and 60

You might be interested in

C) How long is the potato in the air? *hint set h=0*<br> if h=-2(t-3)2+32

Galina-37 [17]

Answer:

if you meant to put $(t-3)^{2}$ , then the answer is 7

Step-by-step explanation:

h=-2 $(t-3)^{2}$ +32

solve for t

0=-2 $(t-3)^{2}$ +32

subtract 32 from both sides

-32=-2 $(t-3)^{2}$

divide both sides by -2

16= $(t-3)^{2}$

find suare root of 16

4= t-3

add 3 to both sides

7 = t

6 0

3 years ago

Consider the equation below.

Korvikt [17]

Answer:

Equation in square form:

$y=3(x+5)^2-4$

Extreme value:

$(h,k)=(-5,-4)$

Step-by-step explanation:

We are given

$y=3x^2+30x+71$

we can complete square

$y=3(x^2+10x)+71$

we can use formula

$a^2+2ab+b^2=(a+b)^2$

$y=3(x^2+2\times x\times 5)+71$

now, we can add and subtract 5^2

$y=3(x^2+2\times x\times 5+5^2-5^2)+71$

$y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71$

$y=3(x+5)^2-75+71$

So, we get equation as

$y=3(x+5)^2-4$

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

$y=a(x-h)^2+k$

where

vertex=(h,k)

now, we can compare and find h and k

$y=3(x+5)^2-4$

we get

h=-5

k=-4

so, extreme value of this equation is

$(h,k)=(-5,-4)$

6 0

3 years ago

Read 2 more answers

Thank you:)))))))))))))

Anastasy [175]

Your welcomeeeeeeeeeee

3 0

2 years ago

Plz help and show me the steps so I know how to do this!! thank you!!

Alex17521 [72]

Answer

6th of the power of 1 would be 5th power

Step-by-step explanation:

3 0

2 years ago

PLEASE HELP!! What are the explicit equation and domain for a geometric sequence with the first term of 3 a second term of -9?

frutty [35]

Answer:

The correct answer is c.

4 0

2 years ago

Read 2 more answers